如何在 NLP 中获得“单词”重要性（TFIDF + Logistic 回归）

Question

我有一个函数可以像这样获得 tfidf 功能：

def get_tfidf_features(data, tfidf_vectorizer=None, ngram_range=(1,2)):
    """ Creates tfidf features and returns them as sparse matrix. If no tfidf_vectorizer is given, 
    the function will train one."""

    if tfidf_vectorizer is not None:
        tfidf = tfidf_vectorizer.transform(data.Comment_text)
    else:
        # only add words to the vocabulary that appear at least 200 times
        tfidf_vectorizer = TfidfVectorizer(min_df=700, ngram_range=ngram_range, stop_words='english')
        tfidf = tfidf_vectorizer.fit_transform(data.Comment_text)        

    tfidf = pd.SparseDataFrame(tfidf.toarray()).to_sparse()
    tfidf.applymap(lambda x: round(x, 4))
    tfidf_features = ['tfidf_' + word for word in tfidf_vectorizer.get_feature_names()]
    tfidf.columns = tfidf_features
    data = data.reset_index().join(tfidf).set_index('index')

    return data, tfidf_vectorizer, tfidf_features    

X_train, tfidf_vectorizer, tfidf_features = get_tfidf_features(X_train)

我应用了这样一个简单的逻辑回归：

logit = LogisticRegression(random_state=0, solver='lbfgs', multi_class='ovr')
logit.fit(X_train.loc[:, features].fillna(0), X_train['Hateful_or_not'])
preds = logit.predict(X_test.loc[:, features].fillna(0))

我得到这样的功能重要性：

 logit.coef_

但这给了我“列”而不是单词的特征重要性

Answer 1

logit.coef_ 确实为您提供了每个单词特征（或二元组）的系数。它将返回一个包含len(features) 元素的数组，其中features 中第i 个位置的单词的系数将位于logit.coef_ 数组的第i 个位置。

例子：

features = ['love','hate','positive','negative']
logit.coef_ = [0.9, -1.2, 0.5, -0.75]

“爱”的系数是0.9，“恨”是-1.2等等……